3.78 \(\int \frac {\text {csch}^6(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=33 \[ -\frac {1}{5} \coth ^5(x)+\frac {\coth ^4(x)}{4}+\frac {\coth ^3(x)}{3}-\frac {\coth ^2(x)}{2} \]

[Out]

-1/2*coth(x)^2+1/3*coth(x)^3+1/4*coth(x)^4-1/5*coth(x)^5

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Rubi [A]  time = 0.05, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3516, 848, 75} \[ -\frac {1}{5} \coth ^5(x)+\frac {\coth ^4(x)}{4}+\frac {\coth ^3(x)}{3}-\frac {\coth ^2(x)}{2} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^6/(1 + Tanh[x]),x]

[Out]

-Coth[x]^2/2 + Coth[x]^3/3 + Coth[x]^4/4 - Coth[x]^5/5

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rubi steps

\begin {align*} \int \frac {\text {csch}^6(x)}{1+\tanh (x)} \, dx &=\operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2}{x^6 (1+x)} \, dx,x,\tanh (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {(-1+x)^2 (1+x)}{x^6} \, dx,x,\tanh (x)\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {1}{x^6}-\frac {1}{x^5}-\frac {1}{x^4}+\frac {1}{x^3}\right ) \, dx,x,\tanh (x)\right )\\ &=-\frac {1}{2} \coth ^2(x)+\frac {\coth ^3(x)}{3}+\frac {\coth ^4(x)}{4}-\frac {\coth ^5(x)}{5}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 27, normalized size = 0.82 \[ \frac {1}{120} \text {csch}^5(x) (30 \sinh (x)-20 \cosh (x)-5 \cosh (3 x)+\cosh (5 x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^6/(1 + Tanh[x]),x]

[Out]

(Csch[x]^5*(-20*Cosh[x] - 5*Cosh[3*x] + Cosh[5*x] + 30*Sinh[x]))/120

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fricas [B]  time = 0.70, size = 185, normalized size = 5.61 \[ -\frac {4 \, {\left (19 \, \cosh \relax (x)^{2} + 42 \, \cosh \relax (x) \sinh \relax (x) + 19 \, \sinh \relax (x)^{2} + 5\right )}}{15 \, {\left (\cosh \relax (x)^{8} + 8 \, \cosh \relax (x) \sinh \relax (x)^{7} + \sinh \relax (x)^{8} + {\left (28 \, \cosh \relax (x)^{2} - 5\right )} \sinh \relax (x)^{6} - 5 \, \cosh \relax (x)^{6} + 2 \, {\left (28 \, \cosh \relax (x)^{3} - 15 \, \cosh \relax (x)\right )} \sinh \relax (x)^{5} + 5 \, {\left (14 \, \cosh \relax (x)^{4} - 15 \, \cosh \relax (x)^{2} + 2\right )} \sinh \relax (x)^{4} + 10 \, \cosh \relax (x)^{4} + 4 \, {\left (14 \, \cosh \relax (x)^{5} - 25 \, \cosh \relax (x)^{3} + 10 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + {\left (28 \, \cosh \relax (x)^{6} - 75 \, \cosh \relax (x)^{4} + 60 \, \cosh \relax (x)^{2} - 11\right )} \sinh \relax (x)^{2} - 11 \, \cosh \relax (x)^{2} + 2 \, {\left (4 \, \cosh \relax (x)^{7} - 15 \, \cosh \relax (x)^{5} + 20 \, \cosh \relax (x)^{3} - 9 \, \cosh \relax (x)\right )} \sinh \relax (x) + 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^6/(1+tanh(x)),x, algorithm="fricas")

[Out]

-4/15*(19*cosh(x)^2 + 42*cosh(x)*sinh(x) + 19*sinh(x)^2 + 5)/(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + (2
8*cosh(x)^2 - 5)*sinh(x)^6 - 5*cosh(x)^6 + 2*(28*cosh(x)^3 - 15*cosh(x))*sinh(x)^5 + 5*(14*cosh(x)^4 - 15*cosh
(x)^2 + 2)*sinh(x)^4 + 10*cosh(x)^4 + 4*(14*cosh(x)^5 - 25*cosh(x)^3 + 10*cosh(x))*sinh(x)^3 + (28*cosh(x)^6 -
75*cosh(x)^4 + 60*cosh(x)^2 - 11)*sinh(x)^2 - 11*cosh(x)^2 + 2*(4*cosh(x)^7 - 15*cosh(x)^5 + 20*cosh(x)^3 - 9
*cosh(x))*sinh(x) + 5)

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giac [A]  time = 0.11, size = 24, normalized size = 0.73 \[ -\frac {4 \, {\left (20 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} - 1\right )}}{15 \, {\left (e^{\left (2 \, x\right )} - 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^6/(1+tanh(x)),x, algorithm="giac")

[Out]

-4/15*(20*e^(4*x) + 5*e^(2*x) - 1)/(e^(2*x) - 1)^5

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maple [B]  time = 0.11, size = 80, normalized size = 2.42 \[ -\frac {\left (\tanh ^{5}\left (\frac {x}{2}\right )\right )}{160}+\frac {\left (\tanh ^{4}\left (\frac {x}{2}\right )\right )}{64}+\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{96}-\frac {\left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{16}+\frac {\tanh \left (\frac {x}{2}\right )}{16}+\frac {1}{96 \tanh \left (\frac {x}{2}\right )^{3}}+\frac {1}{64 \tanh \left (\frac {x}{2}\right )^{4}}+\frac {1}{16 \tanh \left (\frac {x}{2}\right )}-\frac {1}{16 \tanh \left (\frac {x}{2}\right )^{2}}-\frac {1}{160 \tanh \left (\frac {x}{2}\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^6/(1+tanh(x)),x)

[Out]

-1/160*tanh(1/2*x)^5+1/64*tanh(1/2*x)^4+1/96*tanh(1/2*x)^3-1/16*tanh(1/2*x)^2+1/16*tanh(1/2*x)+1/96/tanh(1/2*x
)^3+1/64/tanh(1/2*x)^4+1/16/tanh(1/2*x)-1/16/tanh(1/2*x)^2-1/160/tanh(1/2*x)^5

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maxima [B]  time = 0.31, size = 149, normalized size = 4.52 \[ \frac {4 \, e^{\left (-2 \, x\right )}}{3 \, {\left (5 \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} - 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} - 1\right )}} - \frac {8 \, e^{\left (-4 \, x\right )}}{3 \, {\left (5 \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} - 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} - 1\right )}} + \frac {8 \, e^{\left (-6 \, x\right )}}{5 \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} - 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} - 1} - \frac {4}{15 \, {\left (5 \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} - 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^6/(1+tanh(x)),x, algorithm="maxima")

[Out]

4/3*e^(-2*x)/(5*e^(-2*x) - 10*e^(-4*x) + 10*e^(-6*x) - 5*e^(-8*x) + e^(-10*x) - 1) - 8/3*e^(-4*x)/(5*e^(-2*x)
- 10*e^(-4*x) + 10*e^(-6*x) - 5*e^(-8*x) + e^(-10*x) - 1) + 8*e^(-6*x)/(5*e^(-2*x) - 10*e^(-4*x) + 10*e^(-6*x)
- 5*e^(-8*x) + e^(-10*x) - 1) - 4/15/(5*e^(-2*x) - 10*e^(-4*x) + 10*e^(-6*x) - 5*e^(-8*x) + e^(-10*x) - 1)

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mupad [B]  time = 0.10, size = 24, normalized size = 0.73 \[ -\frac {4\,\left (5\,{\mathrm {e}}^{2\,x}+20\,{\mathrm {e}}^{4\,x}-1\right )}{15\,{\left ({\mathrm {e}}^{2\,x}-1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^6*(tanh(x) + 1)),x)

[Out]

-(4*(5*exp(2*x) + 20*exp(4*x) - 1))/(15*(exp(2*x) - 1)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}^{6}{\relax (x )}}{\tanh {\relax (x )} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**6/(1+tanh(x)),x)

[Out]

Integral(csch(x)**6/(tanh(x) + 1), x)

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